Balancing redox reactions by oxidation number method

There are two very important methods for balancing oxidation-reduction reactions. These are:
  • Oxidation number method
  • Ion-electron method

Oxidation Number Method

During a redox reaction, the total increase in oxidation number must be equal to total decrease in oxidation number. This is the basic principle for balancing chemical equations. In addition, the number of atoms of each kind on one side of the equation must be equal to the number of atoms of the corresponding elements on the other side (the law of conservation of mass should not be violated). The following steps should be followed:

Steps for balancing redox equations by oxidation number method

  • Write the skeleton redox reaction.
  • Indicate the oxidation number of atoms in each compound above the symbol of the element.
  • Identify the element or elements, which undergo a change in oxidation number, one whose oxidation number increases (reducing agent) and the other whose oxidation number decreases (oxidizing agent).
  • Calculate the increase or decrease in oxidation numbers per atom. Multiply this number of increase/decrease of oxidation number, with the number of atoms, which are undergoing change.
  • Equate the increase in oxidation number with decrease in oxidation number on the reactant side by multiplying the formulae of the oxidizing and reducing agents.
  • Balance the equation with respect to all other atoms except hydrogen and oxygen.
  • Finally, balance hydrogen and oxygen.
  • For reactions taking place in acidic solutions, add H+ ions to the side deficient in hydrogen atoms.
  • For reactions taking place in basic solutions, add H2O molecules to the side deficient in hydrogen atoms and simultaneously add equal number to OH- ions on the other side of the equation.
Let us discuss the above method stepwise with the help of reaction between zinc and hydrochloric acid.

Step 1

The skeleton equation is:

steps for balancing redox equations by oxidation number method

Step 2

Oxidation number of various atoms involved in the reaction

Step 3

The oxidation number of zinc has increased from 0 to +2 while that of hydrogen has decreased from +1 to 0. However, the oxidation number of chlorine remains same on both sides of the equation. Therefore, zinc is reducing agent while HCl is oxidizing agent in reaction and the changes are shown as:

Step 4

The increase and decrease in oxidation number per atom can be indicated as: O.N. increases by 2 per atom

Step 5

The increase in oxidation number of 2 per atom can be balanced with decrease in oxidation number of 1 per atom if Zn atoms are multiplied by 1 and HCl by 2. The equation will be:


12. Copper reacts with nitric acid. A brown gas is formed and the solution turns blue. The equation may be written as:

Balance the equation by oxidation number method.


Step 1

Skeleton equation

Step 2

Writing oxidation numbers of each atom

Step 3:

The oxidation number of copper has increased from 0 to +2 while that of nitrogen has decreased from +5 to +4.

Step 4

Show, the increase/decrease of oxidation number

Step 5

Balance the increase/decrease in oxidation number by multiplying NO3- by 2 and Cu by 0.

Step 6

Balance other atoms except H and O as

Step 7

Reaction takes place in acidic medium, so add H+ ions to the side deficient in H+ and balance H and O atoms:


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Vikramvicky Kaler said...

But why cu get multiplied by 0?